Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chapter 3 Stoichiometric Atomic Masses, Mole concept, and Molar Mass (Average atomic.

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

1

Chapter 3Stoichiometric

Atomic Masses, Mole concept, and Molar Mass (Average atomic mass).

Number of atoms per amount of element.Percent composition and Empirical formula

of molecules.Chemical equations, Balancing equations,

and Stoichiometric calculations including limiting reagents.


2

By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams


3

Atomic Masses

Elements occur in nature as mixtures of isotopes

Carbon = 98.89% 12C

1.11% 13C

<0.01% 14C

Carbon atomic mass = 12.01 amu


4

Measuring Atomic Mass

Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer.


5

Spectrum

Most Abundant Isotope


6

Atomic mass of NeonAtomic mass of NeonConsider 100 atoms of neonConsider 100 atoms of neon

18.20100/

2018)

82.8(

)257.0

()92.90(

22

2120

amux

atom

xatoms

xatoms

atomamu

atomamu

atomamu


7

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016100

= 6.941 amu

Average atomic mass of lithium:


8

The Mole

The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

1 mole of anything = 6.022 1023 units of that

thing


9

Avogadro’s number equals

6.022 1023 units


10

Molar Mass

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

C=12 O=16

CO2 = 44.01 grams per mole


11

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)


12

1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu


13

EX. 3.2: Calculate the mass of EX. 3.2: Calculate the mass of 6 atoms of americium in grams6 atoms of americium in grams• From periodic table, Am has a mass of 243

amu.• 6.022x1023 atoms weigh 243 g.

gxatomsAmAmatomsx

g x21

23

1042.2610022.6

243


14

Molar Mass

A chemical compound is a collection of atoms.

What is the mass of 1 mol of CH4?

Mass of 1 mol C = 12.01 g

Mass of 4 mol H = 4x1.008 g

Mass of 1 mol CH4 = 16.04 g


15

Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =


16

S.Ex. 3.7Calculate Molar mass of CaCO3

4.86 moles of CaCO3 → g CaCO3 → g CO32-


17

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%


18

Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen.

Determining Elemental Composition(Formula)


19

The masses obtained (mostly CO2 and H2O and sometimes N2)) will be used to determine:

1. % composition in compound

2. Empirical formula

3. Chemical or molecular formula if the Molar mass of the compound is known or given.


20

Example of Combustion

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O


21

1mol CO244gCO2 1mol C 12g C

22gCO2 mc

mc =

Collect 22.0 g CO2 and 13.5 g H2O

12 16 x 2

44gCO2

22gCO2 x 12gC

% C in CO2 collected

= 6g C


22

1mol C 12g Cnmol C 6g

6g x 1molCnc =

12 gC

= 0.5 mol

Convert g to mole:

Repeat the same for H from H2O

1mol H2O 18g H2O 2 mol H 2g H

13.5g H2O nmol H

2x13.5nmol H = = 1.5 mol H

18 mol H Faster H but still need O

mH


23

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O

2x13.5mH = = 1.5 g H

18mO = 11.5g – mC – mH

= 11.5 – 6 – 1.5 = 4g

m 4nO = = = 0.25 mol O

MM 16


24

OR for C

mC 12.01

1. Fraction of C in CO2 = = =

MMCO2 44.01

2. Mass of C in compound = mass of CO2 x Fraction of C

mC

3. % C in compound = x 100

msample

4. If N then % N = 100 - % C - % H


25

ThenEmpirical Formula

Using the previously calculated % in compound:

% in gram

a. Number of mole of C =

Atomic mass of C

% in gram

b. Number of mole of H =

Atomic mass of H

a b cThen divide by the smallest number:

smallest smallest smallest

: :


26

NoteIf results are : 0.99 : 2.01 : 1.00Then you have to convert to whole numbers:

1 : 2 : 1

CH2N

If results are : 1.49 : 3.01 : 0.99Then you have to multiply by 2:

3 : 6 : 2

C3H6N2

Hence, empirical formula is the simplest formula of a compound


27

S. E.3.1171.65% Cl; 24.27% C ; 4.07 % H

2.021 mol Cl 2.021 mol C 4.04 mol H1 : 1 : 2

Empirical formula is ClCH2

Molar mass = 98.96 g/molMolar mass/ empirical formula mass =

98.96/49.48 = 3

Molecular formula – (ClCH2)2 = Cl2C2H4


28

Formulasmolecular formula = (empirical formula)n

[n = integer]

molecular formula = C6H6 = (CH)6

empirical formula = CH

Then

Molecular Mass

= n

Empirical Mass


29

Figure 3.6: Examples of substances whose empirical and molecular formulas differ. Notice

that molecular formula = (empirical formula)n, where n is a integer.


30

Exercise 3.12

A compound analyzed and found to contain only P and O.

Mass % of P = 43.64%

Molar mass = 283.88 g/mol.

What are the empirical and molecular formulas?


31

Figure 3.8 The Structure of P4O10.


32

Chemical Equations

Chemical change involves a reorganization of the atoms in one or

more substances.


33

Chemical Equation

A representation of a chemical reaction:

C2H5OH + O2 CO2 + H2O

reactants products

Unbalanced !


34

Chemical Equation

C2H5OH + 3O2 2CO2 + 3H2O

The equation is balanced.

1 mole of ethanol reacts with 3 moles of oxygen

to produce

2 moles of carbon dioxide and 3 moles of water


35

Methane Reacts with Oxygen to Produce Flame


36


37

S.Ex. 3.14

Balance the equation

(NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g) + H2O(g)

N and Cr are balanced; 4 for H2O balances H

It also balances O.

(NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g)+4H2O(g)


38

Decomposition of Ammonium Dichromate


39

Decomposition of Ammonium Dichromate


40

S. Ex.3.15Balance the equation

NH3(g) + O2(g)→ NO (g) + H2O (g)

All molecules are of equal complexity.

Balance H first; 2 for NH3 and 3 for H2O

N can be balanced with a coeff. of 2 for NO

O can be balanced by 5/2 for O2

2NH3(g) + 5/2 O2(g)→ 2NO (g) + 3H2O (g)

x2

4NH3(g) + 5O2(g)→ 4NO (g) + 6H2O (g)


41

Calculating Masses of Reactants and Products

1. Balance the equation.

2. Convert mass to moles.

3. Set up mole ratios.

4. Use mole ratios to calculate moles of desired substituent.

5. Convert moles to grams, if necessary.


42

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O


43

2CH3OH + 3O2 2CO2 + 4H2O

2 mol 4 mol

2x(12+4+16) g 4x(2+16) g

209 g m

209 x 4(2+16)m =

2(12+4+16)

OR


44

S. Ex 3.16

What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?

Unbalanced eq.

LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l)

Balanced equation

2 LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l)

23.95 g/mol 44.0 g/mol


45

Limiting Reactant

The limiting reactant is the reactant that is consumed first, limiting the

amounts of products formed.


46

6 green used up6 red left over

Limiting Reagents


47


48

Figure 3.12 Hydrogen and Nitrogen React to Form Ammonia


49

Solving a Stoichiometry Problem

1. Balance the equation.

2. Convert masses to moles.

3. Determine which reactant is limiting.

4. Use moles of limiting reactant and mole ratios to find moles of desired product.

5. Convert from moles to grams.


50

Limiting Reactant

N2 (g) + 3 H2 (g) → 2 NH3 (g)

25.0kg 5.00 kg ??

8.93x102 2.48x103

8.93x102 mol N2 x 3 mol H2 = 2.68x103 mol H2

1 mol N2

H2 is limiting; 2.48x103 mol H2 x 2 mol NH3 = 1.65x103

3 mol H2


51

Limiting Reactant Calculations

What mss of molten iron is produced by 1 kg each of the reactants?

Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)1 mol 2 mol6.26mol 37.04 ??

If all Fe2O3(s) reacts 2x6.26 = 12.5 mol Fe -Limiting

If all Al reacts 37.04 mol Fe -- ExcessThe 6.26 mol Fe2O3 will Disappear first


52

S. Ex 3.182NH3(g) +3CuO(s) → N2(g) +3Cu(s) +3H2O(g)1.06 mol 1.14 mol ??g

If CuO is limiting; 1.14 mol CuO x 1 mol N2

3 mol CuO

= 0.380 mol N2

If NH3 is limiting; 1.06 mol NH3 x 1 mol N2

2 mol NH3

= 0.530 mol N2

Least amount of product obtained with CuO; Cuo is the limiting reactant


53

Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted. Its amount isCalculated using the balanced equation.

Actual Yield is the amount of product actually obtainedfrom a reaction. It is usually given.


54

Percent Yield

Actual yield = quantity of product actually obtained

Theoretical yield = quantity of product predicted by stoichiometry


55

Percent Yield Example

14.4 g

0.104 mol

excess

Actual yield = 6.26 g

1 mol S.A. → 1 mol aspirin0.104 mol S.A. → 0.104 mol aspirin = 18.7 g aspirin

Percent yield = 6.26 g x 100 = 33.5 % 18.7 g


56

Solving a Stoichiometry Problem Invovling Masses of Reactants and Products


57

Sample exercise 3.19 - % yield calculation

Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5Kg CO(g) is reacted with 8.60Kg H 2(g). Calculate the theoretical yield of methanol. If 3.57x104g CH3OH is actually produced, what is the percent yield of methanol?


58

2 H2 (g) + CO(g) → CH3OH (l)

8.60 kg 68.5 kg 35.7 kg(actual yield) 4.27x103mol 2.44x103 mol

If all H2 are converted to methanol,

4.27x103mol H2 x 1 mol methanol = 2.14x103 mol methanol

2 mol H2

If all CO are converted to methanol,2.44x103 mol CO x 1 mol methanol = 2.44x103 mol methanol

1 mol CO H2 is the limiting reactant; theoretical yield = 2.14x103 mol methanol

= 68.6 kg methanol

% yield = actual yield x 100 = 35.7 kg x100 = 52.0%

theoretical yield 68.6 kg


59

Sample Exercise

Titanium tetrachloride, TiCl4, can be made by combining titanium-containing ore (which is often impure TiO2) with carbon and chlorine -

TiO2(s) + 2 Cl2(g) + C(s) TiCl4(l) + CO2(g)

If one begins with 125 g each of Cl2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl4 can be produced?


60

Virtual Laboratory Project


61

Practice Example 1

A compound contains C, H, N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound?

Solution:1. Determine C and H, the rest from 33.5mg is N.2. Determine moles from masses.3. Divide by smallest number of moles.


62

Practice Example 2Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula.

Solution:1. Convert mass to moles.2. Determine empirical formula.3. Determine actual formula.

C8H10N4O2


63

Practice Example 3

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed.


64

Practice Example 5

SnO2(s) + 2 H2(g) Sn(s) + 2 H2O(l)

a) the mass of tin produced from 0.211 moles of hydrogen gas.

b) the number of moles of H2O produced from 339 grams of SnO2.

c) the mass of SnO2 required to produce 39.4 grams of tin.d) the number of atoms of tin produced in the reaction of 3.00

grams of H2.

e) the mass of SnO2 required to produce 1.20 x 1021 molecules of water.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chapter 3 Stoichiometric Atomic Masses, Mole concept, and Molar Mass (Average atomic.

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